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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

Hi bobbym

The formula I got is (n+2)*2^(n-3) for n>=3. The formula there is (n+3)*2^(n-2). We can see that the difference is only in indexing.

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,672

Hi;

You got that formula how, by curve fitting? That only proves for the values you fit for. It does not mean that formula continues for the next diagonal and the one after that.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

We can prove by induction that a(i,j)=2^(i+j-2) for i,j>1. From there, it is easy proving the formula...

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,672

Hi;

What does a(i,j)=2^(i+j-2) for i,j>1 generate?

I would have set it up as

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

Sorry, I got 0 starting arrays in my head.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,672

Hi;

I will leave the inductive proof to you.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

Either way, I think we can be certaing that is how the sequence can be generated...

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,672

Hi;

We can think it all we want. Until we have some proof we ain't gonna convince anybody else of it.

**In mathematics, you don't understand things. You just get used to them.**

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Hi everyone;

I tried to find a formula to obtain the sum of the numbers of each diagonal and this is the result:

(2^n)*2+(2^n)2*n, then I simplified it and I obtained 2^(n-1)*n + 2^(n+1). The result is Number of 1's in all compositions of n+1 (A045623 of OEIS), because 2^(n-1)*n + 2^(n+1. Generate the same terms of (n+3)*2^(n-2), the formula of A045623, proposed by anonimnystefy.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,672

Hi;

To refresh my memory, this square?

1 1 2 4 8

1 1 2 4 8

2 2 4 8 16

4 4 8 16 32

8 8 16 32 64

16 16 32 64 128

**In mathematics, you don't understand things. You just get used to them.**

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Yes, this one.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,672

Hi;

What is holding up some sort of proof for the problem is that my expression given in post #79 does not cover the first row or the first column.

**In mathematics, you don't understand things. You just get used to them.**

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Hi;

So what do you suggest?

*Last edited by Mpmath (2012-11-10 00:11:42)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,672

Hi;

I am working feverishly on an expression that actually generates that table. Then it should be easier to prove the relation is true.

**In mathematics, you don't understand things. You just get used to them.**

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Hi;

Ok. Thanks.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,672

Hi;

Nothing yet.

**In mathematics, you don't understand things. You just get used to them.**

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Hi;

Ok. Keep me informed, thanks.

*Last edited by Mpmath (2012-11-10 11:11:39)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,672

I finally got an expression that will generate the table but it is too complicated for me to understand. At least we can see more of the table.

**In mathematics, you don't understand things. You just get used to them.**

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Hi bobbym;

Good job! Can I see the expression?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,672

Hi;

I did not post it because it is virtually useless.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

That is equal to 2^(i-2)*2^(j-2).

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,672

Aha! You went for the trap. It is incorrect for

It is also incorrect for the whole first column and first row.

**In mathematics, you don't understand things. You just get used to them.**

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**Mpmath****Member**- Registered: 2012-10-11
- Posts: 216

Hi;

Well, good job!

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

bobbym wrote:

Aha! You went for the trap. It is incorrect for

It is also incorrect for the whole first column and first row.

I still think we could use 2^(i-1) and 2^(j-1) for the first column and row respectively, and 2^(i-1) for the rest.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,672

The only problem is that Mathematica gagged on both those series.

I want to sum along the diagonals but if it takes two functions for every diagonal that is going to make the proof much harder or impossible.

**In mathematics, you don't understand things. You just get used to them.**

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